Differentiating inverse trigonometric functions is the key skill here, specifically finding the derivative of arcsec(4x). The formula for the derivative of arcsec(u) where u is a function of x is (1 / (|u| $sqrt(u^2$ - 1))) times u'. In this case, u = 4x, so u' = 4, leading to p'(x) = 4 / (|4x| $sqrt((4x)^2$ - 1)). Simplifying, that's 4 / (|4x| $sqrt(16x^2$ - 1)), which matches choice C. A tempting distractor like choice E fails because it includes a negative sign and uses the wrong form without the absolute value and square root of $u^2$ - 1. Always remember to apply the chain rule when differentiating composite inverse trig functions and double-check the sign based on the specific function.

This problem involves differentiating the inverse tangent function with a linear argument. The derivative of arctan(u) is 1/(1+u²) · u', where u = 5t-1. Using the chain rule, we have d/dt[arctan(5t-1)] = 1/(1+(5t-1)²) · d/dt[5t-1] = 1/(1+(5t-1)²) · 5 = 5/(1+(5t-1)²). Choice D incorrectly forgets to multiply by the derivative of the inner function (which is 5). For inverse trig derivatives, the chain rule factor from the inner function's derivative is essential.

This problem involves differentiating the inverse secant function. The derivative of arcsec(u) is 1/(|u|√(u²-1)) · u', where u = 4x. Using the chain rule, we get d/dx[arcsec(4x)] = 1/(|4x|√((4x)²-1)) · d/dx[4x] = 1/(|4x|√(16x²-1)) · 4 = 4/(|4x|√(16x²-1)). Note that (4x)² = 16x², so √((4x)²-1) = √(16x²-1). Choice C incorrectly omits the chain rule factor of 4. For arcsec and arccsc, remember both the absolute value in the denominator and the chain rule multiplication.

This problem requires differentiating arccsc, another less common inverse trig function. The derivative of arccsc(u) is -u'/(|u|√(u²-1)), where u = x² + 3. Using the chain rule, u' = 2x. Therefore, u'(x) = -2x/(|x²+3|√((x²+3)²-1)), which is choice C. Choice D omits the negative sign essential for arccsc. Choice E lacks the absolute value term. Remember that arccsc, like arccos and arccot, includes a negative sign in its derivative.

This problem requires differentiating arccot with a power function argument. The derivative of arccot(u) is -u'/(1+u²), where u = x³. Using the power rule, u' = 3x². Therefore, q'(x) = -3x²/(1+x⁶), which matches choice B. Choice A omits the negative sign essential for arccot. Choice D incorrectly uses the arccos formula with √(1-u²). Remember that arccot, like arccos, includes a negative sign in its derivative formula.

This problem involves differentiating arcsec, one of the less common inverse trig functions. The derivative of arcsec(u) is u'/(|u|√(u²-1)), where u = 2x + 1. Using the chain rule, u' = 2. Therefore, p'(x) = 2/(|2x+1|√((2x+1)²-1)), which is choice C. Choice A incorrectly uses the arcsin formula, while choice E omits the absolute value. For arcsec and arccsc, always include |u| in the denominator along with √(u²-1).

This problem involves differentiating arccos with a linear composite function. The derivative of arccos(u) is -1/√(1-u²), and we apply the chain rule with u = 1-4s. Differentiating: r'(s) = -1/√(1-(1-4s)²) · d/ds(1-4s) = -1/√(1-(1-4s)²) · (-4) = 4/√(1-(1-4s)²). The two negative signs cancel to give a positive result. Choice B incorrectly keeps the negative sign as -4/√(1-(1-4s)²). For inverse trig derivatives, carefully track signs when both arccos (negative derivative) and decreasing inner functions are involved.

This problem involves differentiating arctangent with a cubic composite function. The derivative of arctan(u) is 1/(1+u²), without any square root, and we apply the chain rule with u = u³. Differentiating: p'(u) = 1/(1+(u³)²) · d/du(u³) = 1/(1+u⁶) · 3u² = 3u²/(1+u⁶). Choice C incorrectly uses the arcsin derivative formula with 1-u² under a square root, giving 3u²/√(1-(u³)²). For inverse trig derivatives, arctangent is unique in having no square root and using 1+u² rather than 1-u² in the denominator.

This problem involves differentiating the inverse sine function, a key skill in AP Calculus BC. The derivative formula for arcsin(u) is u' divided by the square root of 1 minus u squared. Here, u equals x over 2, so u' is 1 over 2. Applying the formula gives (1 over 2) over the square root of 1 minus (x over 2) squared, which is 1 over (2 times the square root of 1 minus (x over 2) squared). A tempting distractor is choice A, which forgets the chain rule factor of 1 over 2 from u'. Always remember to apply the chain rule correctly and verify the sign based on the specific inverse trig function's derivative formula.

This problem involves differentiating the inverse sine function, a key skill in AP Calculus BC. The derivative formula for arcsin(u) is u' divided by the square root of 1 minus u squared. Here, u equals 3x squared, so u' is 6x. Applying the formula gives 6x over the square root of 1 minus (3x squared) squared, which simplifies to 6x over the square root of 1 minus 9x to the fourth. A tempting distractor is choice C, which includes a negative sign, but that would be incorrect because the arcsin derivative formula does not inherently include a negative unlike arccos. Always remember to apply the chain rule correctly and verify the sign based on the specific inverse trig function's derivative formula.